Section 8.6
The following picture is what will be the basis for problems 170-173(it is a slightly modified version of figure 8.14…
For these problems, the process has two steps: (1) draw a picture (which includes shading the area of interest); (2) adding up the probability (or area under the normal curve).
Problem #170
(1) The picture:
(2) Add probabilities: 0.0215 + 0.0013 = 0.0228
Problem #171
(1) The picture:
(2) Add probabilities: 0.0013 + 0.0215 + 0.1359 + 0.3413 + 0.3413 = 0.8413
This can also be done as: 0.5000 + 0.3413= 0.8413 , since the mean to the left is equal to half the area under the curve (0.5000); or the mean divides the curve in half.
Problem #172
(1) The picture:
(2) Add probabilities: 0.3413 + 0.3413 + 0.1359 = 0.8185
Problem #173
(1) The picture:
(2) Add probabilities: 0.1359 + 0.3413 = 0.4772
The following picture is what will be the basis for problems 174-177(it is a slightly modified version of figure 8.14… (the only difference from the earlier picture is in the raw scores)
For these problems, the process has the same two steps: (1) draw a picture (which includes shading the area of interest); (2) adding up the probability (or area under the normal curve).
Problem #174
(1) The picture:
(2) Add probabilities: 0.3413 + 0.3413+ 0.1359 + 0.0215 + 0.0013 = 0.8413
This can also be done as: 0.3413 + 0.5000 = 0.8413 , since the mean to the right is equal to half the area under the curve (0.5000); or the mean divides the curve in half.
Problem #175
(1) The picture:
(2) Add probabilities: 0.0013 + 0.0215 = 0.0228
Problem #176
(1) The picture:
(2) Add probabilities: 0.0215 + 0.1359 = 0.1574
Problem #177
(1) The picture:
(2) Add probabilities: 0.3413 + 0.3413 = 0.6826
The following picture is what will be the basis for problems 178-180 (it is a slightly modified version of figure 8.14… (the only difference from the earlier picture is in the raw scores)
For these problems, the process has the same two steps: (1) draw a picture (which includes shading the area of interest); (2) adding up the probability (or area under the normal curve).
Problem #178
(1) The picture:
(2) Add probabilities: 0.0215 + 0.0013 = 0.0228
Problem #179
(1) The picture:
(2) Add probabilities: 0.0013 + 0.0215 + 0.1359 + 0.3413 + 0.3413 = 0.8413
This can also be done as: 0.5000 + 0.3413= 0.8413 , since the mean to the left is equal to half the area under the curve (0.5000); or the mean divides the curve in half.
Problem #180
(1) The picture:
(2) Add probabilities: 0.1359 + 0.3413 + 0.3413 + 0.1359 = 0.9544
The following picture is what will be the basis for problems 181-183 (it is a slightly modified version of figure 8.14… (the only difference from the earlier picture is in the raw scores)
For these problems, the process has the same two steps: (1) draw a picture (which includes shading the area of interest); (2) adding up the probability (or area under the normal curve).
Problem #181
(1) The picture:
(2) Add probabilities: 0.1359 + 0.0215 + 0.0013 = 0.1587
Problem #182
(1) The picture:
(2) Add probabilities: 0.0013 + 0.0215 = 0.0228
Problem #183
(1) The picture:
(2) Add probabilities: 0.1359 + 0.3413 + 0.3413 + 0.1359 = 0.9544
The following picture is what will be the basis for problems 184-186 (it is a slightly modified version of figure 8.14… (the only difference from the earlier picture is in the raw scores)
For these problems, the process has the same two steps: (1) draw a picture (which includes shading the area of interest); (2) adding up the probability (or area under the normal curve). With the exception of problems 187-188, which require taking the answer from the curve (the above steps) and using it with a binomial.
Problem #184
(1) The picture:
(2) Add probabilities: 0.1359 + 0.0215 + 0.0013 = 0.1587
Problem #185
(1) The picture:
(2) Add probabilities: 0.0013 + 0.0215 = 0.0228
Problem #186
(1) The picture:
(2) Add probabilities: 0.3413 + 0.1359 = 0.9544
Problems 187 and 188 are a combination of normal and binomial…you know that it is binomial because of the phrases “at least 3 of the next 5” or “at most 4 of the next 5”…and from these phrases we know what n (number of trials) is…FIVE. The p (probability of success) is based on the phrase “faster than 70”; and since 70 is the mean it means that the p = 0.5…now onto the specifics:
Problem #187
“at least 3 of the next 5” means 3, 4 and 5, so we need the probabilities of each and add them together (in some way). There are three methods as you may recall, the first is to calculate each:
P(3) = C(5,3)*0.53*0.52 = 0.3125
P(4) = C(5,3)*0.53*0.52 = 0.15625
P(5) = C(5,3)*0.53*0.52 = 0.03125
Answer = P(3) + P(4) + P(5) = 0.3125+0.15625+0.03125 = 0.5
Or the second way is to use the table and look up the sum from 0 to 2 (the part you do not want) and subtract from 1…looking in the section for n=5 trials and under the column for a p of 0.5…you get 0.500, so 1 – 0.5 = 0.5
And the third way is similar to the second except that you use the TI calculator to compute the value of the sum from 0 to 2, using binomialcdf(5,0.5, 2) which gives you 0.5, so 1 – 0.5 = 0.5
Problem #188
“atmost4 of the next 5” means 4 and 5, so we need the probabilities of each and add them together (in some way). There are three methods as you may recall, the first is to calculate each:
P(4) = C(5,3)*0.53*0.52 = 0.15625
P(5) = C(5,3)*0.53*0.52 = 0.03125
Answer = P(4) + P(5) = 0.15625+0.03125 = 0.1875
Or the second way is to use the table and look up the sum from 0 to 3 (the part you do not want) and subtract from 1…looking in the section for n=5 trials and under the column for a p of 0.5…you get 0.812, so 1 – 0.812 = 0.188
And the third way is similar to the second except that you use the TI calculator to compute the value of the sum from 0 to 3, using binomialcdf(5,0.5, 3) which gives you 0.8125, so 1 – 0.8125 = 0.1875